14 I ELEMENTS OF THE THEORY OF SETS Let ji-*•<*, be a bijection of N onto A; then n-+f(an) is a mapping of N onto B, and we can therefore suppose A = N. For each b e B, f~l(b) is not empty by assumption; let m(b) be its smallest element. Then/(/n(6)) = *, which shows at once that m is an injective mapping of B into N; m can be considered as a bijection of B onto m(B) <= N, and by (1.9.1) m(S) is at most denumerable. Q.E.D. We observe that if a set A is at most denumerable, there is always a surjection of N onto A; this is obvious if A is infinite; if not, there is a bijection /of an interval Q^f^m onto A, and one extends/to a surjection by putting g(n) =/(w) for n > m. (1.9.3) The set N x N = N2 is denumerable. We define an injection / of N x N into N by putting ("diagonal enumeration"; it turns out to be a bijection, but we do not need that result). Indeed, ifx + y=*a, then (a +l)(a + 2)/2 = a + I + a(a + l)/2; hence if x + y < x' + /, as y^a, f(x, y)^a + a(a + l)/2 /A(w) of N onto AA . Let A = (j AA , and consider the mapping (m, n) -»/An(w) of A 6 1^ N x N into A; this mapping is surjective, for if x e AM , there is an n such that 11 = ln, and an m such that x =/M(w) =f^n(m). The result now follows from (1.9.3) and (1.9.2) since A is infinite. The result (1.9.4) is still valid if the word "denumerable" is everywhere replaced by "at most denumerable." We have only to replace bijections by surjections in the proof, using the remark which follows (1 .9.2). Finally, we consider the following result as an axiom: (1 .9.5) Every infinite set contains a denumerable subset. s finite